I've recently been studying i.i.d random variables and I think I'm starting to grasp the concept. I've been set this question but been given very little guidance on how to answer it and am very confused. Any help is greatly appreciated.
Consider i.i.d. random variables $(X_i)$ drawn from a uniform distribution on $[0, 1].$
In the following find scaling sequences $a_n$, $b_n$ such that $a_n(M_n − b_n)$ converges in distribution to a non-trivial limit function $G$.
(a) $Y_i = X_i,$ and $M_n = \max(Y_1, . . . , Y_n);$
(b) $U_i = \frac{1}{X_i}$, and $M_n = \max(U_1, . . . , U_n)$;
In each case find first of all the probability distribution function $P(M_n ≤ u/a_n + b_n)$ as a function of $u_n = u/a_n + b_n$. Then find suitable scaling sequences $a_n$, $b_n$ so that you get a non-trivial limit $G(u)$ as $n → ∞$.
Let me try to set you on your way. First of all, try to determine the distribution of $M_n$. Notice that
$$P(M_n\leq x) = P(X_i\leq x, \forall i)$$
This plus the independence and equality in distribution of the $X_i$ should lead you to the following formula:
$$P(M_n\leq x) = \begin{cases}0 &, \text{for} \; x<0 \; ,\\ x^n &, \text{for} \; 0\leq x<1 \; ,\\ 1 &, \text{for} \; 1\leq x \; .\end{cases}$$
Replace $x$ by $\frac{u}{a_n}+b_n$. You'll see that the middle term should be a familiar form, it should remind you of
$$\lim_{n\to \infty}\left(1+\frac{u}{n}\right)^n = e^u \; .$$
EDIT For part b) you'll obtain
$$P\left(M_n\leq \frac{u}{a_n}+b_n\right) = \begin{cases}0 &, \text{for} \; \frac{u}{a_n}+b_n\leq 1 \; ,\\ \left(1-\frac{1}{\frac{u}{a_n}+b_n}\right)^n &, \text{for} \; \frac{u}{a_n}+b_n > 1 \; .\end{cases}$$
Putting $a_n=\frac{1}{n}$ and $b_n=n$ this becomes
$$P(M_n\leq n(u+1)) = \begin{cases}0 &, \text{for} \; n(u+1)\leq 1 \; ,\\ (1-\frac{1}{n(u+1)})^n &, \text{for} \; n(u+1)> 1 \; .\end{cases}$$
The limiting case is then
$$\lim_{n\to +\infty}P\left(\frac{M_n}{n}-1\leq u\right) = \begin{cases}0 &, \text{for} \; u \leq -1 \; ,\\ e^{-\frac{1}{u+1}} &, \text{for} \; u>-1 \; .\end{cases}$$